TITLE: Vector $x$ Solves Least Squares Problem $\iff$ $b - Ax \in \text{Null}(A^*)$
QUESTION [6 upvotes]: I am preparing for a graduate exam in Numerical Linear Algebra and I have to solve the following question (without calculus, and without any other theorems about Least Squares):

Let $A \in \mathbb{C}^{m \times n}$ with $n < m$ (not necessarily of full rank). For any $x \in \mathbb{C}^n$ let $r(x) = b - Ax$. Show that $x$ solves the least squares problem $\min_{x \in \mathbb{R}^n}||b - Ax||_2$ if and only if $r(x) \in \text{Null}(A^*).$

Below, I wrote down my attempt. Not only is it incomplete (it purports to prove just one direction of the equivalence), but later I realized that it contains a major flaw. I would immensely appreciate your help  in either fixing this solution, or finding a new one.
I have very little time on the exam, so as a bonus, I would be very thankful if you could tell me if you see bad habits in my mathematical writing which make the solution too long.
P.S. I have an inkling that a correct solution might involve a QR decomposition, but I'm not sure.

Flawed Solution:
Lemma: The equation $A^*Ap = A^*b$ has a solution.
Proof: We are trying to show that $A^*b \in \text{Col}(A^*A) = \text{Null}(A^*A)^{\perp}$. Take any $y \in \text{Null}(A^*A).$ We will show that $A^*b \perp y$.
But first, we will show $Ay = 0$. Indeed, $$y \in \text{Null}(A^*A) \implies A^*Ay = 0 \implies y^*A^*Ay = 0 \implies ||Ay|| = 0 \implies Ay = 0.$$
Now, we show perpendicularity. We have
$$(A^*b)^*y = b^*(Ay) = b^*0 = 0.$$
So $A^*b \in \text{Col}(A^*A)$.
Q.E.D.

We proceed to the main proof. Let $p$ be a solution of $A^*Ap = A^*b$. Then $A^*b - A^*Ap = 0$.
We claim $r(p) \perp \text{Col(A)}.$ Indeed, for any $Az \in \text{Col}(A)$, we have
$$(Az)^*(b - Ap) = z^*(A^*b - A^*Ap ) = 0.$$
Now, by the Pythagorean Theorem,
$$||b - Ay||^2 = ||b - Ap + Ap - Ay||^2 = ||b - Ap||^2 + ||Ap - Ay||^2 \ge ||b - Ap||^2$$
with equality $\iff Ay = Ap$.
$\color{red}{\text{The flaw is here. I need to bound the function } ||b - Ax||}$
$\color{red}{\text{ by a constant, not by a function of p.}}$
Now suppose $x$ is a vector that solves the L.S. problem. Then $Ax = Ay$, so $r(x) = b - Ax = b - Ap = r(p)$
and $r(p) \in \text{Null}(A^*)$ because $A^*Ap = A^*b \implies A^*(b - Ap) = 0$.
Now for the converse, suppose that $A^*Ax = A^*b$. I don't yet know how to show that $Ax = Ap$.

REPLY [4 votes]: I'll use the fact that $N(A^*)$ is the orthogonal complement of $R(A)$. (I like to call this fact the "four subspaces theorem" because it shows how the four subspaces $R(A), N(A), R(A^*)$, and $N(A^*)$ are related.)
There is a unique way to decompose $b$ as
$$
\tag{1}b = b_1 + b_2
$$
with $b_1 \in R(A)$ and $b_2 \in N(A^*)$. Notice that if $x \in \mathbb C^n$ then $b_1 - Ax \in R(A)$, and so $b_1 - Ax \perp b_2$. It follows that
\begin{align}
\|b - Ax \|_2^2 &= \|b_2 + b_1 - Ax \|_2^2 \\
&= \| b_2 \|^2 + \| b_1 - Ax \|_2^2.
\end{align}
This expression is minimized when $Ax = b_1$, or equivalently when $b - Ax = b_2$.
So, if $x$ minimizes $\| r(x) \|_2$, then $r(x) \in N(A^*)$. On the other hand, suppose that $r(x) \in N(A^*)$. Comparing the equation
$$
b = Ax + r(x)
$$
with the unique decomposition (1) reveals that $r(x) = b_2$, which means that $\| r(x) \|_2$ is minimal.

REPLY [3 votes]: Note that $at^2+bt+c$ has a minimum  for $t=0$ iff $b=0$ and $a\geq 0$.
You may then consider minimizing $\|r(x)\|^2=(Ax-b,Ax-b)$. Replacing $x$ by $x+ty$ with $t$ real and $y$ a vector yields:
$$ \|r(x+ty)\|^2 =
            \|r(x)\|^2 + 2t (y,A^* r(x)) + t^2(Ay,Ay)$$
By the above, $\|r(x)\|^2$ is the minimum value iff for every $y$ the linear term vanishes, or equivalently, iff $A^* r(x)=0$.